Sequences and Series Test

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Sequences and Series Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Value of $$9 + 99 + 999 + .... $$ upto $$n$$ terms is

A
$$\displaystyle \frac{10^{n}-9n-10}{9}$$

B
$$\displaystyle \frac{10^{n}-9n-10}{81}$$

C
$$\displaystyle \frac{10^{n+1}-9n-10}{81}$$

D
$$\displaystyle \frac{10^{n+1}-9n-10}{9}$$

Solution

Given, $$9 + 99 + 999 + .... $$ upto $$ n $$ terms
we can write this series in this form
   $$s=(10-1)+(100-1)+(1000-1).......$$ up to $$n$$ terms.
    $$= [(10+100+1000+......)+(-1-1-1-.......\text {up to n terms})]$$
    $$= [(10+10^2+10^3+.....\text {up to n terms})-n]$$                ........(i)
sum of n terms of GP. $$=\dfrac {a(r^n-1)}{r-1}$$, where $$a$$ is first term and $$r$$ is the common ratio of the GP.
$$\text {We have the GP } 10+10^2+10^3+.... \text {up to n terms} $$, here first term $$a=10$$ and common ratio $$r=10$$
So,$$10+10^2+10^3+...... \text {n terms}= \dfrac {10(10^n-1)}{10-1} = \dfrac {10^{n+1}-10}{9}$$On Substituting$$s=10+10^2+10^3+...... \text {n terms}= \dfrac {10^{n+1}-10}{9}$$ in equation (i). we get,
    $$ =\dfrac {10^{n+1}-10}{9}-n = \dfrac {10^{n+1}-9n-10}{9} $$
$$\Rightarrow 9+99+999+.......\text {up to n terms} = \dfrac {10^{n+1}-9n-10}{9} $$
Option D is correct.
Question 2
1 / -0

The sum of the series $$4 + 8 + 16 + 32 + .......$$. till $$10$$ terms is

A
$$6\, \times\, 2^{10}\, -\, 1$$

B
$$4\, \times\, 2^{10}\, +\, 1$$

C
$$4(2^{10}\, -\, 1)$$

D
$$4(2^{10}\, +\, 1)$$

Solution

Let $$S_{10}=4 + 8 + 16 + 32 + ........ $$ upto $$10 $$ terms
$$=4(1+2+4+8+.........$$.upto $$10$$ terms
Clearly, $$1+2+2^2+2^3+...$$ is an GP with first term $$=1$$ and common ratio $$=2$$.
$$\therefore S_{10}=4[\displaystyle \frac{1(2^{10}-1)}{2-1}]=4(2^{10}-1)$$
Thus the sum of the given series is $$4(2^{10}-1)$$
Question 3
1 / -0

$$\displaystyle \sum _{ p=1 }^{ 32 }{ \left( 3p+2 \right) { \left[ \sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi }{ 11 } } -i\cos { \frac { 2q\pi }{ 11 } } \right) } \right] }^{ p } } =$$

A
$$8\left( 1-i \right) $$

B
$$16\left( 1-i \right) $$

C
$$48\left( 1-i \right) $$

D
None of these

Solution

$$\displaystyle\sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi }{ 11 } -i } \sin { \frac { 2q\pi }{ 11 } } \right) } $$
$$\displaystyle=\left\{ \sin { \frac { 2\pi }{ 11 } } +\sin { \frac { 4\pi }{ 11 } } +...+10 \right\}- i\left\{ \cos { \frac { 2\pi }{ 11 } } +\cos { \frac { 4\pi }{ 11 } } +...+10 \right\} $$
$$\displaystyle=\frac { \sin { \left( \frac { 2\pi }{ 11 } +\frac { 9\pi }{ 11 } \right) } \sin { \frac { 10\pi }{ 11 } } }{ \sin { \frac { \pi }{ 11 } } } -i\frac { \cos { \left( \frac { 2\pi }{ 11 } +\frac { 9\pi }{ 11 } \right) \sin { \frac { 10\pi }{ 11 } } } }{ \sin { \frac { \pi }{ 11 } } } $$$$=0-i\left( -1 \right) =i$$ ...(1)
$$\displaystyle\therefore S=\sum _{ p=1 }^{ 32 }{ \left( 3q+2 \right) { \left[ \sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi }{ 11 } -i } \cos { \frac { 2q\pi }{ 11 } } \right) } \right] }^{ p } } $$
$$\displaystyle=\sum _{ p=1 }^{ 32 }{ \left( 3p+2 \right) { i }^{ p } } =3\sum _{ p=1 }^{ 32 }{ p } { i }^{ p }+2\sum _{ p=1 }^{ 32 }{ { i }^{ p } } =3A+2B$$ ...(2)
Now, $$\displaystyle A=i+2{ i }^{ 2 }+3{ i }^{ 3 }+...+32{ i }^{ 32 }$$$$\displaystyle \Rightarrow Ai={ i }^{ 2 }+2{ i }^{ 3 }+...+31{ i }^{ 32 }+32{ i }^{ 33 }$$
$$\displaystyle\Rightarrow A\left( 1-i \right) =i+{ i }^{ 2 }+...+{ i }^{ 32 }-32{ i }^{ 33 }$$$$\displaystyle =\frac { { i }^{ 32 }-1 }{ i-1 } -32i=-32i\quad \left[ \because { i }^{ 32 }=1 \right] $$
$$\displaystyle\therefore A=\frac { -32i }{ 1-i } =\frac { 32\left( 1+i \right) }{ 2 } =16\left( 1-i \right) $$ ...(3)
and, $$\displaystyle B=i+{ i }^{ 2 }+...+{ i }^{ 32 }=0$$ ...(4)
Hence, $$\displaystyle S=3\times 16\left( 1-i \right) =48\left( 1-i \right) .$$
Question 4
1 / -0

Find the sum the infinite G.P.:
$$1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}$$

A
$$\displaystyle \frac{3}{5}$$

B
$$\displaystyle \frac{3}{2}$$

C
$$\displaystyle \frac{49}{27}$$

D
$$\displaystyle \frac{8}{5}$$

Solution

Formula used: Sum of infinite GP
$$S_\infty=a+ar+ar^2+ar^3+...=\dfrac{a}{1-r}$$, where $$|r|<1$$
Given, $$1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}$$
Clearly, it is a GP with first term $$a=1$$ and common difference $$d=\frac13$$
$$\therefore1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}=\frac{1}{1-\frac13}=\frac32$$
Hence, option B is correct.
Question 5
1 / -0

Evaluate $$7 + 77 + 777 + .............$$ upto $$n$$ terms.

A
$$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n]$$

B
$$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n\, -\, 10]$$

C
$$\displaystyle \frac{7}{81}[10^n\, -\, 9n\, -\, 10]$$

D
$$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, n\, -\, 10]$$

Solution

Let $$S=7+77+777+........+ $$ up to $$n$$ terms
$$\Rightarrow \dfrac{7}{9}\left [ 9+99+999+..... \right ]$$

$$\Rightarrow \dfrac{7}{9}\left [\left (10-1 \right )+\left (10^2-1 \right )+\left (10^3+1 \right )+........up\quad to\quad n\quad terms \right ] $$

$$\Rightarrow \dfrac{7}{9}\left [10+10^2+10^3+......+10^n-n \right ]$$

$$\Rightarrow \dfrac{7}{9}\left [ 10\frac{10^{n}-1}{9}-n \right ]$$

$$\Rightarrow \dfrac{7}{81}\left [ 10^{n+1}-9n-10 \right ]$$
Question 6
1 / -0

For $$\frac {2^2+4^2+6^2+...+(2n)^2}{1^2+3^2+5^2+...+(2n-1)^2}$$ to exceed 1.01, the maximum value of n is

A
99

B
100

C
101

D
150

Solution

$$\frac {2^2[1^2+2^2+...+n^2]}{[1^2+3^2+5^2+...+(2n-1)^2]}$$
$$1^2+2^2+3^2+....+(2n)^2=\frac {2n(2n+1)(4n+1)}{6}$$
$$[1^2+3^2....+(2n-1)^2+2^2[1^2+2^n.....+n^2]$$
$$=\frac {2n(2n+1)(4n+1)}{6}$$
$$S+4\frac {n(n+1)(2nn+1)}{6}=\frac {2n(2n+1)(4n+1)}{6}$$
$$S+\frac {2n(2n+1)(4n+1)}{6}-\frac {4n(n+1)(2n+1)}{6}$$
$$=2n\left (\frac {2n+1}{6}\right )[4n+1-2n-2]$$
$$=\frac {2n(2n+1)(2n-1)}{6}$$
$$Ratio =\frac {4n(n+1)(2n+1)}{6}\times \frac {6}{2n(2n+1)(2n-1)}=\frac {2n+2}{2n-1}$$
$$\frac {2n+2}{2n-1} > \frac {101}{100}$$
$$200n+200 > 202 n-101$$
$$2n < 301$$
$$n < \frac {301}{2}\Rightarrow $$ maximum value $$=150$$
Question 7
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle.Choose the correct answer for the term marked by the symbol (?)

A
$$18$$

B
$$40$$

C
$$21$$

D
$$24$$

Solution
Question 8
1 / -0

The sum of 'n' terms of series $$1^2+(1^2+2^2)+(1^2+ 2^2+ 3^2)+(1^2+2^2+ 3^2+4^2)+ .......$$ will be

A
$$\displaystyle \frac{1}{4} n (n + 1)^2 (n + 2)$$

B
$$\displaystyle \frac{1}{3} n (n + 1)^2 (n + 2)$$

C
$$\displaystyle \frac{1}{12} n (n + 1) (n + 2)^2$$

D
$$\displaystyle \frac{1}{12} n (n + 1)^2 (n + 2)$$

Solution

The given series is $${ 1 }^{ 2 }+({ 1 }^{ 2 }+2^{ 2 })+({ 1 }^{ 2 }+2^{ 2 }+3^{ 2 })+({ 1 }^{ 2 }+2^{ 2 }+3^{ 2 }+4^{ 2 })+.....$$

$${ a }_{ n }=({ 1 }^{ 2 }+2^{ 2 }+3^{ 2 }+.......+n^{ 2 })$$

$$=\dfrac { n(n+1)(2n+1) }{ 6 } \\ =\dfrac { n(2n^{ 2 }+3n+1) }{ 6 } \\ =\dfrac { 2n^{ 3 }+3n^{ 2 }+n }{ 6 } \\ =\dfrac { 1 }{ 3 } n^{ 3 }+\dfrac { 1 }{ 2 } n^{ 2 }+\dfrac { 1 }{ 6 } n$$

Therefore, we have:

$${ S }_{ n }=\sum _{ k=1 }^{ n }{ { a }_{ k } } \\ =\sum _{ k=1 }^{ n }{ \left( \dfrac { 1 }{ 3 } k^{ 3 }+\dfrac { 1 }{ 2 } k^{ 2 }+\dfrac { 1 }{ 6 } k \right) } \\ =\dfrac { 1 }{ 3 } \sum _{ k=1 }^{ n }{ k^{ 3 } } +\dfrac { 1 }{ 2 } \sum _{ k=1 }^{ n }{ k^{ 2 } } +\dfrac { 1 }{ 6 } \sum _{ k=1 }^{ n }{ k } \\ =\dfrac { 1 }{ 3 } \times \dfrac { n^{ 2 }+(n+1)^{ 2 } }{ 2^{ 2 } } +\dfrac { 1 }{ 2 } \times \dfrac { n(n+1)(2n+1) }{ 6 } +\dfrac { 1 }{ 6 } \times \dfrac { n(n+1) }{ 2 }$$
$$=\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n(n+1) }{ 2 } +\dfrac { (2n+1) }{ 2 } +\dfrac { 1 }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n^{ 2 }+n+2n+1+1 }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n(n+1)+2(n+1) }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { (n+1)(n+2) }{ 2 } \right] \\ =\dfrac { n(n+1)^{ 2 }(n+2) }{ 12 }$$

Hence, the sum is $$\dfrac { n(n+1)^{ 2 }(n+2) }{ 12 }$$.
Question 9
1 / -0

$$11, 26, 56, 101, 161, .....$$

A
$$236$$

B
$$226$$

C
$$216$$

D
$$206$$

Solution

Add $$15, 30, 45, 60, 75$$ to each number Next number is $$161 + 75 = 236$$
Question 10
1 / -0

The value of $$\displaystyle\frac{1}{1\cdot2\cdot3}+\displaystyle\frac{1}{2\cdot3\cdot4}+\displaystyle\frac{1}{3\cdot4\cdot5}+\displaystyle\frac{1}{4\cdot5\cdot6}$$ is equal to

A
$$\displaystyle\frac{7}{30}$$

B
$$\displaystyle\frac{11}{30}$$

C
$$\displaystyle\frac{13}{30}$$

D
$$\displaystyle\frac{17}{30}$$

Solution

Given $$\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}$$

$$\Rightarrow \dfrac{1}{1\times 2\times3 }+\dfrac{1}{2\times 3\times 4}+\dfrac{1}{3\times 4\times 5}+\dfrac{1}{4\times 5\times 6} $$

$$\Rightarrow \dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+\dfrac{1}{120}$$

$$= \dfrac{20+5+2+1}{120}$$

$$= \dfrac{28}{120}$$

$$= \dfrac{7}{30}$$

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Sequences and Series Test - 23

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Sequences and Series Test - 23

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SHARING IS CARING

Question 1

$$\displaystyle \frac{10^{n}-9n-10}{9}$$

$$\displaystyle \frac{10^{n}-9n-10}{81}$$

$$\displaystyle \frac{10^{n+1}-9n-10}{81}$$

$$\displaystyle \frac{10^{n+1}-9n-10}{9}$$

Solution

Question 2

$$6\, \times\, 2^{10}\, -\, 1$$

$$4\, \times\, 2^{10}\, +\, 1$$

$$4(2^{10}\, -\, 1)$$

$$4(2^{10}\, +\, 1)$$

Solution

Question 3

$$8\left( 1-i \right) $$

$$16\left( 1-i \right) $$

$$48\left( 1-i \right) $$

None of these

Solution

Question 4

$$\displaystyle \frac{3}{5}$$

$$\displaystyle \frac{3}{2}$$

$$\displaystyle \frac{49}{27}$$

$$\displaystyle \frac{8}{5}$$

Solution

Question 5

$$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n]$$

$$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n\, -\, 10]$$

$$\displaystyle \frac{7}{81}[10^n\, -\, 9n\, -\, 10]$$

$$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, n\, -\, 10]$$

Solution

Question 6

99

100

101

150

Solution

Question 7

$$18$$

$$40$$

$$21$$

$$24$$

Solution

Question 8

$$\displaystyle \frac{1}{4} n (n + 1)^2 (n + 2)$$

$$\displaystyle \frac{1}{3} n (n + 1)^2 (n + 2)$$

$$\displaystyle \frac{1}{12} n (n + 1) (n + 2)^2$$

$$\displaystyle \frac{1}{12} n (n + 1)^2 (n + 2)$$

Solution

Question 9

$$236$$

$$226$$

$$216$$

$$206$$

Solution

Question 10

$$\displaystyle\frac{7}{30}$$

$$\displaystyle\frac{11}{30}$$

$$\displaystyle\frac{13}{30}$$

$$\displaystyle\frac{17}{30}$$

Solution

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